RVHS Prelim P2_Soln

1 River Valley High School 9729/02 [Turn over 2018 Preliminary Examinations MARK SCHEME 1 (a) W: Si SiCl4 + 2H2O → SiO2 + 4HCl X: P PCl5 + CH3COOH → POCl3 + CH3COCl OR 3CH3COOH + PCl3 ⎯→ 3CH3COCl + H3PO3 [4] (b) (i) Al2O3(s) + 2NaOH(aq) → 2Na[Al(OH)4](aq) + H2O(l) OR Al2O3(s) + 2OH−(aq) → 2[Al(OH)4]−(aq) + H2O(l) [3] (b) (ii) On addition of HC l(aq), a white ppt is formed. Ppt would dissolve in excess HCl(aq) to give a colourless solution. (c) BaCO3 has a higher thermal decomposition temperature. Ba2+ has a larger ionic radius compared to Z2+/Mg2+. Ba2+ hence has a lower charge density, and polarise the CO 32– anion to a lower extent (compared to Z2+/Mg2+), and the C–O bonds are weakened to a smaller extent, resulting in a higher thermal decomposition temperature. (d) The 3p electron to be removed from A l is at a higher energy level compared to the 3s electron to be removed fr om Mg, hence the p electron is less strongly attracted to the nucleus and require less energy to remove. [1] [Total: 9]

2 River Valley High School 9729/02 [Turn over 2018 Preliminary Examinations 2 (a) For pH = 3.5 [H+] = 10–3.5 = 3.16 x 10–4 moldm–3 Let the number of moles of C5H5NHCl be Y mol Since pKa = 5.25 , Ka = 10–5.25 = 5.62 × 10–6 (3.16 × 10–4)2 / Y – (3.16 × 10–4) = 5.62 × 10–6 Y = [(3.16 × 10–4)2 / 5.62 × 10–6 ] + (3.16 × 10–4) = 0.0181 mol Or (3.16 × 10–4)2 / Y = 5.62 × 10–6 Y = [(3.16 × 10–4)2 / 5.62 × 10–6 ] = 0.0178 mol Mass to be added = (0.0181 × (12.0 × 5 + 1.0 × 6 + 14.0 + 35.5)) ÷ 0.98 = 2.13 g (or 2.10 g) [3] (b) [C5H5N] = [(0.100 × 0.025) – (0.0125 × 0.005)] / 0.03 = 0.0813 mol dm–3 [pyridinium chloride] = (0.0125 × 0.005) / 0.03 = 2.08 × 10–3 mol dm–3 pKb =14 – pKa = 8.75 pOH = pKb + lg [salt]/[base] pOH = 8.75 + lg (2.08 × 10–3 / 0.0813) = 7.16 pH = 14 – 7.16 = 6.84 [3] (c) On addition of a small amount of acid (H +) to the buffer solution, nearly all the added H+ ions are neutralised by the large amount of C5H5N. Hence [H+] does not increase appreciably and the pH is kept approximately constant. C5H5N (aq) + H+(aq) → C5H5NH+(aq) On addition of a small amount of base (OH −) to the buffer solution, nearly all the added OH− ions are neutralised by the large amount of C5H5NH+. Hence [OH−] does not increase appreciably and the pH is kept approximately constant. C5H5NH+ (aq) + OH−(aq) → C5H5NH (aq) + H2O(l) [3] [Total: 9]

3 River Valley High School 9729/02 [Turn over 2018 Preliminary Examinations 3 (a) Initial P of NO(g) = ଵ ସ × 5 = 1.25 atm Initial P of NO2Cl(g) = 5 – 1.25 = 3.75 atm NO2Cl + NO ⇌ NOCl + NO2 Initial P / atm 3.75 1.25 0 0 Δ in P / atm –0.85 –0.85 +0.85 +0.85 Eqm P / atm 2.90 0.40 0.85 0.85 Kp = ሺ଴.଼ହሻሺ଴.଼ହሻ ሺଶ.ଽ଴ሻሺ଴.ସ଴ሻ = 0.623 (no units) [2] (b) [2] (c) [1] (d) Electrophilic addition [3] [Total: 8] PNO2Cl / atm time / min 0 t1 t2 t3 t4 2.90 3.25 3.75 O N Cl ××